A) \[\sqrt{2}\]
B) \[\frac{1}{\sqrt{2}}\]
C) \[1\]
D) \[\left( \sqrt{2}+1 \right)\]
Correct Answer: B
Solution :
\[\tan A=\sqrt{2}-1\] \[\sin 2A=\frac{2\tan A}{1+{{\tan }^{2}}A}=\frac{2\left( \sqrt{2}-1 \right)}{1+{{\left( \sqrt{2}-2 \right)}^{2}}}\] \[=\frac{2\left( \sqrt{2}-1 \right)}{\left( 1+2+1-2\sqrt{2} \right)}=\frac{2\left( \sqrt{2}-1 \right)}{2\left( 2-\sqrt{2} \right)}\] \[=\frac{\left( \sqrt{2}-1 \right)\left( 2+\sqrt{2} \right)}{\left( 2-\sqrt{2} \right)\left( 2+\sqrt{2} \right)}=\frac{2\sqrt{2}-2+2-\sqrt{2}}{\left( 4-2 \right)}\] \[=\frac{\sqrt{2}}{2}=\underline{\frac{\mathbf{1}}{\sqrt{\mathbf{2}}}}\]
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